A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is 200 mm 2 . To keep a metal at 727 ° C , heat energy flowing through this hole per sec, in joules, is ( take σ = 5 .67 × 10 - 8 W m - 2 K - 4 )

Question

A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is 200 mm 2 . To keep a metal at 727 ° C , heat energy flowing through this hole per sec, in joules, is ( take σ = 5 .67 × 10 - 8 W m - 2 K - 4 ), 22 . 68, 2 . 268, 1 . 134, 11 . 34

MARKS App · Accepted Answer

By Stefan's law, if the absolute temperature of a black body is T , then the energy E emitted per second per unit surface area of the body E ∝ T 4 E = σ A T 4 Given, σ = 5 .67 × 10 - 8 W m - 2 K - 4 T = 727 + 273 = 1000 K A = 200 × 10 - 6 m E = 5.67 × 10 - 8 × 200 × 10 - 6 × 1000 4 = 5.67 × 2 × 10 - 12 × 10 12 = 11 .34 J s - 1

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